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The Bennequin bound [1] says that, for a transverse knot (or later link) $K$ in $S^3$, $$\mathrm{sl}(K) \le - \chi(\Sigma)$$ for any Seifert surface $\Sigma$ for $K$, where $\mathrm{sl}$ is the self-linking number.

Lisca and Matic proved [2, Theorem 3.4] that for a Legendrian knot $K$ $$|r(K)| + \mathrm{tb}(K) \le 2g(\Sigma) - 1$$ for any smooth surface $\Sigma \subset B^4$ with boundary $K$, where $\mathrm{tb}$ and $r$ are the Thurston-Bennequin and rotation numbers.

EDIT: This was actually proved by Rudolph [3].

I have several questions here.

(a) Why is the Lisca-Matic bound always stated in terms of Legendrian knots? The combination $|r| + \mathrm{tb}$ is exactly the self-linking number of the transverse push-off, with one of the two orientations. Wouldn't it be simpler to just state it in terms of transverse knots?

(b) Isn't the Lisca-Matic bound strictly stronger than the Bennequin bound? Why isn't it more popular to quote it? Maybe it's about the extension to links (which, in the case of the Bennequin bound, is due to Eliashberg)? It seems that Lisca and Matic only state the result for knots in their paper, but it seems to me that the proof should carry over.

For knots/links in more general 3-manifolds, the Bennequin bound applies as long as you have a tight contact structure, while to state Lisca-Matic you need some sort of filling.

[1] Bennequin, "Entrelacements et équations de Pfaff", In Third Schnepfenried Geometry Conference, Vol. 1, Schnepfenried, 1982, 87–161. Astérisque 107. Paris: Société Mathématique de France, 1983.

[2] Lisca, P., and G. Matić. “Stein 4-manifolds with Boundary and Contact Structures.” 55–66. Topology and Its Applications 88, nos. 1–2, 1998.

[3] Rudolph, Lee, "Quasipositivity as an obstruction to sliceness." Bull. Amer. Math. Soc. (N.S.) 29 (1993), no. 1, 51–59.

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To answer the question in the title: yes, the slice Bennequin inequality is stronger than the Bennequin inequality. For instance, it shows that all slice knots have negative maximal self-linking number, and I can't see how to prove this from Bennequin alone. (Just for the record, there are strengthening of both inequalities due to Olga Plamenevskaya, where she replaces the slice genus by Ozsváth–Szabó's $\tau$ and by Rasmussen's $-s/2$.)

What follows is probably not the answer to the question(s) you're asking, but I think it's still relevant. (I think that especially (b) came from an original misunderstanding of your question on my part; I'll still leave it there for reference.)

(a) You're right that it would be simpler to state it in terms of self-linking numbers, but —the way I see it— the inequality arises more naturally from the Legendrian viewpoint. The proof I have in mind starts with a Legendrian knot $K$ and attaches a Weinstein 2-handle along it, to get a 2-handlebody that I'll denote with $X_L$; this handlebody has a complex structure $J_L$.

In this way the two classical invariants appear naturally: the Thurston–Bennequin number as the framing (well, actually, ${\rm tb}\,L-1$, and not ${\rm tb}\,L$) and the rotation number as the evaluation of the first Chern class $c_1(J_L)$ of the complex structure $J_L$ on the generator of $H_2(X_L)$. So, at least from this perspective, the two invariants arise as separate components of the adjunction formula, and "pedagogically" it makes more sense to state it in terms of $\rm tb$ and $\rm rot$ rather than in terms of the self-linking number of the push-off.

Off hand, I don't see a corresponding transverse proof. Of course you can use transverse invariants in knot Floer homology, and that's less biased towards Legendrian than the previous proof, but I think it's also less geometric. Or you can use Plamenevskaya's invariant in Khovanov homology (which is only transverse), but, as above, this is somewhat less direct/geometric.

(b) On the face of it, yes, the inequality with $|{\rm rot}|$ is stronger than the one without absolute values. But you can always reverse the orientation of $L$, and this changes the sign of $\rm rot$ while keeping $\rm tb$ fixed, so the self-linking inequalities for $L$ and $-L$ give you the same amount of information.

There is also a less obvious, but more contact-flavoured, construction, namely the Legendrian mirror. This construction shows that the "mountain range" picture for $({\rm tb}, {\rm rot})$ is symmetric: given $L$ you can produce the Legendrian mirror of $L$, $\mu(L)$ which has $({\rm tb}, {\rm rot})(\mu(L)) = ({\rm tb}, -{\rm rot})(L)$. In a front projection, if I recall correctly, this is just rotation by 180 degrees.


As a side note, I think that the Heegaard Floer (or monopole) proof of the slice Bennequin inequality only requires that the contact structure has non-trivial (Ozsváth–Szabó, say) contact invariant, and it's stronger than Bennequin's inequality, in the sense that it should give a bound on the "slice genus" in $Y\times I$.

I would expect that Matt Hedden wrote something about this in his Notions of positivity and the Ozsváth–Szabó concordance invariant paper.

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    $\begingroup$ Thank you, that's very helpful. Before going on to the fancier stuff, the first thing I was reminded of is that I should be calling this the slice-Bennequin inequality, and attributing it to Rudolph, who proved it some years before the Lisca-Matic paper. You sketch the Lisca-Matic proof, which indeed uses both components. Rudolph gives a different proof, not directly connecting it to either transverse or Legendrian knots, instead working with braids (which he uses to write the RHS). I still feel there should be an strictly transverse proof. $\endgroup$ – Dylan Thurston Jul 18 '18 at 16:34
  • $\begingroup$ I was aware that Rudolph was the first to prove it, but I'm not familiar with his proof. If it's a braid proof, though, it smells a lot more like a transverse proof than a Legendrian one, since closure of braids can be naturally viewed as transverse knots (and vice-versa), but this is not the case for Legendrian knots. $\endgroup$ – Marco Golla Jul 18 '18 at 21:13
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    $\begingroup$ The grid diagrams picture really clarified the relationship between Legendrian, and transverse knots and braids for me: arxiv.org/abs/0812.3665 $\endgroup$ – Dylan Thurston Jul 19 '18 at 2:57
  • $\begingroup$ (That is, both Legendrian and braids give transverse knots, but a Legendrian knot does not give a braid nor vice versa, as you alluded to.) $\endgroup$ – Dylan Thurston Jul 19 '18 at 2:58
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Yes, the slice-Bennequin bound is stronger than Bennequin bound:

(a) For example, the knot $K \# - K$ is ribbon, so the slice-Bennequin implies that any transverse representative must have negative self-linking number. But the Bennequin bound only implies it is less than $4 g(K) - 1$, which is positive if $K$ is nontrivial.

(b) Given any transverse representatives $T, -T$ of a knot and its mirror, we can take a transverse connected sum to obtain a knot satisfying $sl(T \# - T) = sl(T) + sl(T) + 1 \leq -1$. Let $T$ be the positive torus knot $T(p,q)$. This maximizes the Bennequin bound so we can assume $sl(T) = pq - p - q$. But this implies that the self-linking number of any transverse representative of the negative torus knot $T(p,q)$ is as most $p + q - pq - 2$.

As Marco points out, one reason to state the result in terms of Legendrian knots is because we have a good local model for attaching a 2-handle along a Legendrian knot. There is a theory for attaching handles along transverse knots, described in David Gay's thesis, but it is trickier. The approach of Lisca-Matic is to use the genus bound coming from the global adjunction inequality in a Kahler surface of general-type to deduce a slice-genus bound.

Finally, complex points and Lai indices give another, very satisfying conceptual relation between the Bennequin-Eliashberg bound and the Adjunction Inequality that is more "transverse" than "Legendrian". A generic real, immersed surface $\Sigma$ in a complex surface $X$ will have finitely many complex points --- i.e. points where the tangent plane is a complex line. These can be either positive or negative (according to whether the orientation of $\Sigma$ agrees with the complex orientation) and either elliptic or hyperbolic. The Lai indices of $\Sigma$ count these points with signs:

$I_+(\Sigma) = e_+ - h_+ \qquad I_-(\Sigma) = e_- - h_-$

If $\Sigma$ is embedded, these satisfy the relations

$I_+(\Sigma) + I_-(\Sigma) = \chi(\Sigma) + [\Sigma]^2$

$I_+(\Sigma) - I_-(\Sigma) = \langle c_1(X),[\Sigma] \rangle$

The adjunction inequality is then equivalent to the inequality $I_-(\Sigma) \leq 0$.

There is an analogous story for a convex surface $\Sigma$ in a contact manifold $(M,\xi)$. The tangencies of $\xi$ along $\Sigma$ are the singularities of the characteristic foliation. These can be either positive or negative and either elliptic or hyperbolic. We can then define Lai indices exactly as above. If $\Sigma$ is a Seifert surface bounded by a transverse knot $K$, then

$sl(K) = I_-(\Sigma) - I_+(\Sigma)$

Moreover, the Bennequin-Eliashberg bound is equivalent to the inequality $I_-(\Sigma) \leq 0$. Eliashberg proves this directly by showing how to perturb the surface to eliminate all negative elliptic points.

See the books Stein manifolds and holomorphic mappings by Forstneric and Surgery on contact 3-manifolds and Stein surfaces by Ozbagci-Stipsicz.

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